地下水水文學--4

• 1. Groundwater flow to wells(井流)
• • 1.1 Steady-flow: Thesis equation
  • • 1.1.1 steady radial flow in confined aquifer due to pumping--Thesis equation
    • 1.1.2 steady radial flow in unconfined aquifer due to pumping
  • 1.2 Transient flow to well--unsteady-state(confined aquifer--Theis solution)

2022.04.02

1. Groundwater flow to wells(井流)

1.1 Steady-flow: Thesis equation

landfill填埋場
Terminology related well pumping

---

$h_0$, static water level

$s(t)$, drawdown, 降深

$specificcapacity,S_c$,

fully penetrating well
完整井，貫通整個含水層的井，非完整井，反之

1.1.1 steady radial flow in confined aquifer due to pumping–Thesis equation

$M,K,r_w,h_0,Q,s(r),h(r)$
2-D, homogenous, isotropic, steady-state, uniform

Firstly, for confined aquifer
$$KM\frac{{?}^{2}h}{?x^2}+KM\frac{{?}^{2}h}{?y^2}=0$$

for well flow

$$KM\frac{{?}^{2}h}{?x^2}+KM\frac{{?}^{2}h}{?y^2}+W(x,y)=0$$

柱坐標

$$x=rcos(\theta ),y=rsin(\theta ),x^2+y^2=r^2$$

將直角坐標轉換為柱坐標，帶入方程，則G.E.
$$KM\frac{1}{r}\frac{d}{dr}(r\frac{dh}{dr})=0,井流的基本方程$$

boundary condition

$$\begin{gathered} h=h_0,r=R \\ Q=K2\pi rM\frac{?h}{?r},r=r_w, \\ 流量邊界,流入井的流量，側面流進去，最大高度只能是M \end{gathered}$$
在任意位置
$$Q=K2\pi rM\frac{?h}{?r},r=r_1$$
Q(r1)與抽水量Q的關系，$Q(r_1)=Q$

$$\begin{gathered} Q=K2\pi rM\frac{?h}{?r} \\ \int Q/rdr=\int KM2\pi dh \\ h=\frac{Q}{2\pi T}ln(r)+C \\ {h}_0=\frac{Q}{2\pi T}ln(R)+C \end{gathered}$$
得
$$s=h_0?h=\frac{Q}{2\pi T}ln(\frac{R}{r})$$
寫成log
$$\begin{gathered} s=h_0?h=\frac{Q}{2.73T}lg(\frac{R}{r}),降深公式 \\ Theimequationforconfinedaquiferduetowell \end{gathered}$$

則兩個觀測井，可以觀測到兩個及兩個降深$s_1,s_2$
$$s_1?s_2=\frac{Q}{2\pi T}ln(\frac{r_2}{r_1})$$

基于以上公式，可以進行參數估計，求得導水系數$T$, 抽水試驗確定導水系數。只能確定滲透系數或導水系數。

1.1.2 steady radial flow in unconfined aquifer due to pumping

和承壓不一樣在于有自由水面

2-D, isotropic, with Dupuit assumption
$K,r_w,h_0,Q,s(r),h(r)$
G.E.
$$\begin{gathered} K\frac{1}{r}\frac{d(h\frac{dh}{dr})}{dr}=0 \\ h在里面非線性 \end{gathered}$$

Boundary conditon
$$Q=KAi=K2\pi rh\frac{dh}{dr}$$
推導
$$\int hdh=\int \frac{Q}{2\pi Kr}dr$$
最終得；
$$\begin{gathered} h_0^2?h^2=\frac{Q}{\pi K}ln(\frac{R}{r}), \\ Dupuitwellflowequation \end{gathered}$$
變形：

$$\begin{gathered} (h_0?h)(h_0+h)=\frac{Q}{\pi K}ln(\frac{R}{r}) \\ | \\ s(2h_0?s)=\frac{Q}{\pi K}ln(\frac{R}{r})=\frac{Q}{1,36K}lg(\frac{R}{r}) \end{gathered}$$
參數估計：已知降深，可以求$K$

以上均為理想狀態。we need transient

1.2 Transient flow to well–unsteady-state(confined aquifer–Theis solution)

求解方程，考慮水位隨時間變化
homo, isotropic, confined aquifer, $h(r,t)$

$$T\frac{1}{r}(r\frac{?h}{?r})=S\frac{?h}{?t}$$
展開為：
$$\frac{{?}^{2}h}{?r^2}+\frac{1}{r}\frac{?h}{?r}=\frac{S}{T}\frac{?h}{?r}$$
寫成降深得形式，帶入$s=h_0?h$
$$\begin{gathered} \frac{{?}^{2}s}{?r^2}+\frac{1}{r}\frac{?s}{?r}=\frac{S}{T}\frac{?s}{?r} \\ s(r,t)=0,t=0 \\ s(r,t)=0,r?>+\infty \\ \underset{r?>0}{\lim }r\frac{?s}{?r}=\frac{?Q}{2\pi T},r=0 \end{gathered}$$

$$Q=?2\pi Tr\frac{?s}{?r}$$
井處的流量最大，離井越遠，流量越小
$$Q_i<Q$$

Solution
$$s(r,t)=\frac{Q}{4\pi T}W(u),Theisequaiton$$

$$\begin{gathered} W(u)={\int }_u^{\infty }\frac{e^{?x}}{x}dx \\ u=\frac{r^{2}s}{4Tt} \end{gathered}$$
W(u)為泰斯井函數

部分內容轉載自南方科技大學梁修雨教授地下水水文學講課PPT

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